Chapter A.3, Problem 61E

Solution

Simplify.

  $\frac{1}{3-x}$ .

Given:

The given fraction is $\frac{3}{x^2+3x}-\frac{1}{x}-\frac{6}{x^2-9}$ .

Concept Used:

• Remove any grouping symbol such as brackets and parentheses by multiplying factors.
• Use the exponent rule to remove grouping if the terms are containing exponents.
• Combine the like terms by addition or subtraction.
• Combine the constants.

Calculation:

The given fraction is $\frac{3}{x^2+3x}-\frac{1}{x}-\frac{6}{x^2-9}$ .

  $\begin{array}{c}\frac{3}{x^2+3x}-\frac{1}{x}-\frac{6}{x^2-9}=\frac{3}{x(x+3)}-\frac{1}{x}-\frac{6}{x^2-3^2}\\ =\frac{3}{x(x+3)}-\frac{1}{x}-\frac{6}{(x+3)(x-3)}\\ =\frac{3(x-3)-(x+3)(x-3)-6x}{x(x+3)(x-3)}\\ =\frac{3x-9-(x^2-3^2)-6x}{x(x+3)(x-3)}\\ =\frac{3x-9-x^2+3^2-6x}{x(x+3)(x-3)}\\ =\frac{3x-6x-9+9-x^2}{x(x+3)(x-3)}\\ =\frac{-3x-x^2}{x(x+3)(x-3)}\\ =\frac{-x(3+x)}{x(x+3)(x-3)}\\ =\frac{-1}{(x-3)}\\ =\frac{1}{3-x}\end{array}$

Thus by simplifying $\frac{3}{x^2+3x}-\frac{1}{x}-\frac{6}{x^2-9}$ to a single fraction it comes $\frac{1}{3-x}$ .